# Abstract Nonsense

## Left Regular Representation (Pt. II)

Point of post: This post a is a continuation of this one.

Left Multiplication Representation

With the above inner product we can now successfully describe the canonical $\ast$-representation of $\mathcal{A}\left(G\right)$ into $\text{End}\left(\mathcal{A}\left(G\right)\right)$ endowed with the canonical inner product. Namely:

Theorem: Let $G$ be a finite group and $\mathcal{A}\left(G\right)$ be the group algebra on $G$. Then, if $\mathcal{A}\left(G\right)$ is given the canonical inner product then the map $\varrho:\mathcal{A}\left(G\right)\to\text{End}\left(\mathcal{A}\left(G\right)\right)$ given by $\varrho_a(b)=a\ast b$ is a $\ast$-representation.

Proof: First of all it’s worth showing that the map $\varrho_a:\mathcal{A}\left(G\right)\to\mathcal{A}\left(G\right)$ really does define a linear transformation. But, this is fairly trivial to check since for any $\beta,\gamma\in\mathbb{C}$ and $a,b,c\in\mathcal{A}\left(G\right)$

\begin{aligned}\varrho_a\left(\beta b+\gamma c\right) &= a\ast\left(\beta b+\gamma c\right)\\ &= a\ast\left(\beta b\right)+a\ast\left(\gamma c\right)\\ &= \beta\left(a\ast b\right)+\gamma\left(a\ast c\right)\\ &= \beta\varrho_a(b)+\gamma\varrho_a(c)\end{aligned}

It thus remains to show that $\varrho$ is a $\ast$-representation of $\mathcal{A}\left(G\right)$. To see that $\varrho$ is additive we merely note that for any $a,b,c\in\mathcal{A}\left(G\right)$ we have that

$\displaystyle \varrho_{a+b}(c)=\left(a+b\right)\ast c=a\ast c+b\ast c=\varrho_a(c)+\varrho_b(c)$

from where it follows from the arbitrariness of $c$ that $\varrho_{a+b}=\varrho_a+\varrho_b$. Similarly one has that

$\displaystyle \varrho_{a\ast b}(c)=(a\ast b)\ast c=a\ast (b\ast c)=\varrho_a(b\ast c)=\varrho_a\varrho_b(c)$

and thus by the arbitrariness of $c$ it follows that $\varrho_{a\ast b}=\varrho_a\varrho_b$. Noticing that evidently for any $a\in\mathcal{A}\left(G\right)$ one has that $\varrho_{\delta_e}(a)=\delta_e\ast a=a$ it’s clear that $\varrho_{\delta_e}=\mathbf{1}$. Thus, it remains show that that $\displaystyle \varrho_{a^\ast}=\varrho_a^\ast$. To do this we recall that $\varrho_a^\ast$ is the unique endomorphism on $\mathcal{A}\left(G\right)$ such that

$\left\langle \rho_a(b),c\right\rangle=\left\langle b,\rho_a^\ast(c)\right\rangle$

for all $b,c\in\mathcal{A}\left(G\right)$.  Thus, it suffices to show that $\rho_{a^\ast}$ satisfies this property. Indeed, let $b,c\in\mathcal{A}\left(G\right)$ and note that

\displaystyle \begin{aligned}\left\langle \rho_a(b),c\right\rangle &= \left\langle a\ast b,c\right\rangle\\ &= \frac{1}{|G|}\sum_{g\in G}(a\ast b)(g)\overline{c(g)}\\ &= \frac{1}{|G|}\sum_{g\in G}\sum_{k\in G}a\left(gk^{-1}\right) b(k)\overline{c(g)}\\ &= \frac{1}{|G|}\sum_{g\in G}\sum_{k\in G}\overline{\overline{a\left(\left(kg^{-1}\right)^{-1}\right)}}b(k)\overline{c(g)}\\ &= \frac{1}{|G|}\sum_{g\in G}\sum_{k\in G}\overline{a^{\ast}\left(kg^{-1}\right)}b(k)\overline{c(g)}\\ &= \frac{1}{|G|}\sum_{k\in G}\left(\sum_{g\in G}\overline{a^\ast(kg^{-1})c(g)}\right)b(k)\\ &= \frac{1}{|G|}\sum_{k\in G}b(k)\overline{\left(a^\ast\ast c\right)(k)}\\ &= \left\langle b,a^\ast \ast c\right\rangle\\ &= \left\langle b,\rho_{a^{\ast}}(c)\right\rangle\end{aligned}

from where by the arbitrariness of $a,b\in\mathcal{A}\left(G\right)$ we are able to conclude, by the aforementioned characterization of the adjoint, that $\rho_{a^\ast}=\rho_a^\ast$. The theorem then follows. $\blacksquare$

Left Regular Representation

We have just seen that the map $\varrho:\mathcal{A}\left(G\right)\to\text{End}\left(\mathcal{A}\left(G\right)\right)$ given by $\varrho_a(b)=a\ast b$ is a $\ast$-representation on $\mathscr{A}\left(G\right)$. But, by previous theorem we know that this induces a representation $\rho:G\to\mathcal{U}\left(\mathcal{A}\left(G\right)\right)$ given by $g\mapsto \varrho(\delta_g)$. At first this may seem like quite an intractable definition since, in practice, $\varrho(\delta_g)$ means nothing. What’s interesting is that $\varrho(\delta_g)$ acts on elements of $\mathcal{A}\left(G\right)$ in a particularly simple way. In particular, $\left(\rho_g(a)\right)(h)=a\left(g^{-1}h\right)$. Indeed:

\displaystyle \begin{aligned}\left(\rho_g(a)\right)(h) &= \left(\varrho_{\delta_g}(a)\right)(h)\\ &= \left(\delta_g\ast a\right)(h)\\ &= \sum_{k\in G}\delta_g\left(hk^{-1}\right)a(k)\\ &= \delta_g\left(h\left(g^{-1}h\right)^{-1})\right)a\left(g^{-1}h\right)\\ &= a\left(g^{-1}h\right)\end{aligned}

This representation is called the left regular representation of $G$ on $\mathcal{A}\left(G\right)$.

Now, while this is an enlightening way to look at the left regular representation there is an admittedly easier method to prove that it’s a representation. Namely, it’s clear that if $\rho$ is the left regular representation then $\rho$ is a homomorphism $G\to\text{End}\left(\mathcal{A}\left(G\right)\right)$ and thus it would suffice to check that $\rho_g$ is unitary for each $g\in G$. Note though that $\rho_g(\delta_h)=\delta_{gh}$ from where it clearly follows that $\rho_g$ takes the orthonormal basis $\{\sqrt{|G|}\delta_h\}_{h\in G}$ to itself, and thus (by a common characterization of unitarity) each $\rho_g$ is unitary.

We shall see that the left regular representation will play a decisive role in the theory to come.

References:

1.Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print

January 23, 2011 -

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